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Math Help - Projectile motion - maximum height

  1. #1
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    Projectile motion - maximum height

    If a projectile is fired upward from the surface of the moon with an initial velocity of 40 m/s, its height, in metres, after t seconds is given by s=40t-.83t^2
    a) Estimate the instantaneous velocity when t = a, (without using calculus)
    b) Using your answer in part a, determine the maximum height.

    Appreciate any help.
    Last edited by mr fantastic; June 5th 2009 at 04:40 PM.
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  2. #2
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    Quote Originally Posted by peekay View Post
    If a projectile is fired upward from the surface of the moon with an initial velocity of 40 m/s, its height, in metres, after t seconds is given by s=40t-.83t^2
    a) Estimate the instantaneous velocity when t = a, (without using calculus)
    b) Using your answer in part a, determine the maximum height.

    Appreciate any help.
    I'd start by calculating the average velocity between the times t = a and t = a + h and then take the limit as h --> 0.
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  3. #3
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    I don't know what the variable a and h are referring to ,

    There were actually two questions before question a and b, and they were to find the average velocity between 1 and 3 seconds which is 38.705 m/s and to also find the instantaneous velocity at 2 seconds, which is 38.4763 m/s. Am I suppose to do something with the instantaneous velocity when t = 2?
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  4. #4
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    Talking

    In the earlier parts, they gave you a = 1 and a + h = 3 (so h = 2). Here you have a and a + h, so do the exact same steps as you did before.

    The only difference will be that you won't be able to simplify as much, because you're working with a variable instead of which numbers.
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  5. #5
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    So

    a + h = (40h-.38h^2) + (40a-.38a^2)
    a = (40a-.38a^2)

    and then just do m = (y2-y1) / (x2-x1) ?
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  6. #6
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    Could anyone solve this and then explain step by step how they did it?

    Still having trouble with it
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  7. #7
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    Quote Originally Posted by peekay View Post
    So

    a + h = (40h-.38h^2) + (40a-.38a^2)
    a = (40a-.38a^2)

    and then just do m = (y2-y1) / (x2-x1) ?
    The line in red is wrong. You have the function

    S(t) = 40t - 0.83t^2.

    (I wanted to use function notation so I added the parentheses and the t.) You plug in a + h for t:

    \begin{aligned}<br />
S(a + h) &= 40(a + h) - 0.83(a + h)^2 \\<br />
&= 40a + 40h -0.83(a^2 + 2ah + h^2) \\<br />
&= 40a + 40h -0.83a^2 -1.66ah - 0.83h^2 \\<br />
&= -0.83a^2 - 0.83h^2 - 1.66ah + 40a + 40h<br />
\end{aligned}

    Also, the line underneath the red one above isn't completely correct. You're not really saying that a equals (40a-.38a^2), but this is what the expression gives you when you plug in a. In function notation,

    S(a) = 40a - 0.83a^2.

    Now use the formula for the slope. The two points are (a, S(a)) and (a + h, S(a + h)):

    \begin{aligned}<br />
m &= \frac{y_2 - y_1}{x_2 - x_1} \\<br />
&= \frac{S(a + h) - S(a)}{(a + h) - a} \\<br />
&= \frac{(-0.83a^2 - 0.83h^2 - 1.66ah + 40a + 40h) - (40a - 0.83a^2)}{h} \\<br />
&= \frac{-0.83h^2 - 1.66ah + 40h}{h} \\<br />
&= \frac{h(-0.83h - 1.66a + 40)}{h}<br />
\end{aligned}
    m = -0.83h - 1.66a + 40

    Now take the limit as h -> 0:
    \lim_{h \to 0} (-0.83h - 1.66a + 40) = -1.66a + 40

    For part b, set -1.66a + 40 equal to 0 and solve for a.


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